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\title{\heiti\zihao{2} 习题15.2}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{利用复合函数求导法则,求下列偏导数}
\subsection{$z=u^{2} \ln v, u=\dfrac{x}{y}, v=3 x-2 y,$ 求 $\dfrac{\partial z}{\partial x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =\dfrac{\partial z}{\partial u}\cdot\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\cdot\dfrac{\partial v}{\partial x} \\
		                               & =\dfrac{2u\ln v}{y}+\dfrac{3u^2}{v}                                                                                                  \\
		                               & =\dfrac{2x\ln(3x-2y)}{y^2}+\dfrac{3x^2}{y^2(3x-2y)}
	\end{aligned}
$$

\subsection{$u=\dfrac{y}{x}, x=\mathrm{e}^{t}, y=1-\mathrm{e}^{2 t},$ 求 $\dfrac{\mathrm{d} u}{\mathrm{~d} t}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\mathrm{~d} u}{\mathrm{~d} t} & = \dfrac{\mathrm{~d} u}{\mathrm{~d} x}\cdot\dfrac{\mathrm{~d} x}{\mathrm{~d} t}+\dfrac{\mathrm{~d} u}{\mathrm{~d} y}\cdot\dfrac{\mathrm{~d} y}{\mathrm{~d} t} \\
		                                     & =-\dfrac{y\mathrm{e}^t}{x^2}-\dfrac{2\mathrm{e}^{2t}}{x}                                                                                                      \\
		                                     & =-(\mathrm{e}^t+\mathrm{e}^{-t})
	\end{aligned}
$$

\subsection{$z=\mathrm{e}^{u} \sin v, u=x y, v=x+y,$ 求 $\dfrac{\partial z}{\partial y}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial y} & =\dfrac{\partial z}{\partial u}\cdot\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\cdot\dfrac{\partial v}{\partial y} \\
		                               & =x\mathrm{e}^u\sin v+\mathrm{e}^u\cos v                                                                                              \\
		                               & =x\mathrm{e}^{xy}\sin(x+y)+e^{xy}\cos(x+y)
	\end{aligned}
$$

\subsection{$u=f\left(x y, \dfrac{x}{y}\right),$ 求 $\dfrac{\partial u}{\partial x}, \dfrac{\partial u}{\partial y}$}
\textbf{解}\quad
记$\boldsymbol{f}(\boldsymbol{x})=\left(\begin{array}{c}
			f^1(\boldsymbol{x}) \\
			f^2(\boldsymbol{x}) \\
			\vdots              \\
			f^n(\boldsymbol{x}) \\
		\end{array}\right)$,则记$f_{i} = \dfrac{\partial \boldsymbol{f}}{\partial f^i}$,从而
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x} & = f_1\cdot \dfrac{\partial f^1}{\partial x}+f_2\cdot\dfrac{\partial f^2}{\partial x} \\
		                               & =f_1\cdot y+f_2\cdot\dfrac{1}{y}                                                     \\
		\dfrac{\partial u}{\partial y} & = f_1\cdot \dfrac{\partial f^1}{\partial y}+f_2\cdot\dfrac{\partial f^2}{\partial y} \\
		                               & =f_1\cdot x-f_2\cdot\dfrac{x}{y^2}                                                   \\
	\end{aligned}
$$

\subsection{$z=h(u, x, y), y=g(u, v, x), x=f(u, v);$求$ \dfrac{\partial z}{\partial u}, \dfrac{\partial z}{\partial v}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial u} & =h_1+h_2\cdot\dfrac{\partial x}{\partial u}+h_3\cdot\dfrac{\partial y}{\partial u} \\
		                               & =h_1+h_2\cdot f_1+h_3\cdot (g_1+g_3\cdot\dfrac{\partial x}{\partial u})            \\
		                               & =h_1+h_2\cdot f_1+h_3\cdot g_1+h_3\cdot g_3\cdot f_1                               \\
		\dfrac{\partial z}{\partial v} & =h_2\cdot\dfrac{\partial x}{\partial v}+h_3\cdot\dfrac{\partial y}{\partial v}     \\
		                               & =h_2\cdot f_2+h_3\cdot (g_2+g_3\cdot\dfrac{\partial x}{\partial v})                \\
		                               & =h_2\cdot f_2+h_3\cdot g_2+h_3\cdot g_3\cdot f_2
	\end{aligned}
$$

\subsection{$z=f(u, x, y), x=g(v, w), y=h(u, v) ;$ 求 $\dfrac{\partial z}{\partial u}, \dfrac{\partial z}{\partial v}, \dfrac{\partial z}{\partial w}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial u} & =f_1+f_3\cdot y_1          \\
		\dfrac{\partial z}{\partial v} & =f_2\cdot g_1+f_3\cdot h_2 \\
		\dfrac{\partial z}{\partial w} & =f_2\cdot g_2
	\end{aligned}
$$

\subsection{$w=f(x, y, z), x=u+v, y=u-v, z=u v ;$ 求 $\dfrac{\partial w}{\partial u}, \dfrac{\partial w}{\partial v}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{\partial w}{\partial u} & = f_1\cdot \dfrac{\partial x}{\partial u}+f_2\cdot\dfrac{\partial y}{\partial u}+f_3\cdot\dfrac{\partial z}{\partial u} \\
		                               & =f_1 + f_2 + f_3\cdot v                                                                                                 \\
		\dfrac{\partial w}{\partial v} & =f_1\cdot \dfrac{\partial x}{\partial v}+f_2\cdot\dfrac{\partial y}{\partial v}+f_3\cdot\dfrac{\partial z}{\partial v}  \\
		                               & =f_1-f_2+f_3\cdot u
	\end{aligned}
$$

\section{利用链式法则求下列复合映射 $f \circ g$ 在指定点的 $Jacobi$ 矩阵}
\subsection{$f(x, y)=\left(x, y, x^{2} y\right), g(s, t)=\left(s+t, s^{2}-t^{2}\right)$, 在点 $(s, t)=(2,1)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\boldsymbol{f}\circ\boldsymbol{g}    & = \left(\begin{array}{c}
				f_1(g_1,g_2) \\
				f_2(g_1,g_2) \\
				f_3(g_1,g_2)
			\end{array}\right)                                             \\
		(\boldsymbol{f}\circ\boldsymbol{g})' & =\left(\begin{array}{c}
				f_1'(\boldsymbol{g}) \\
				f_2'(\boldsymbol{g}) \\
				f_3'(\boldsymbol{g})
			\end{array}\right)\cdot \boldsymbol{g}'                        \\
		                                     & =\left(\begin{array}{c}
				\nabla f_1(\boldsymbol{g}) \\
				\nabla f_2(\boldsymbol{g}) \\
				\nabla f_3(\boldsymbol{g})
			\end{array}
		\right)\cdot\left(\begin{array}{cc}
				\dfrac{\partial g_1}{\partial s} & \dfrac{\partial g_1}{\partial t} \\
				\dfrac{\partial g_2}{\partial s} & \dfrac{\partial g_2}{\partial t}
			\end{array}\right)                                                                         \\
		                                     & =\left(\begin{array}{cc}
				1   & 0   \\
				0   & 1   \\
				2xy & x^2
			\end{array}\right)\cdot\left(\begin{array}{cc}
				1  & 1   \\
				2s & -2t
			\end{array}\right) \\
		                                     & =\left(\begin{array}{cc}
				1                           & 1                           \\
				2s                          & -2t                         \\
				2(s+t)(s^2-t^2) + 2s(s+t)^2 & 2(s+t)(s^2-t^2) - 2t(s+t)^2
			\end{array}\right)                                             \\
	\end{aligned}
$$
将$(s,t)=(2,1)$带入即得
$$
	(\boldsymbol{f}\circ\boldsymbol{g})'=\left(\begin{array}{cc}
			1  & 1  \\
			4  & -2 \\
			54 & 0
		\end{array}\right)
$$

\subsection{$f(x, y, z)=\left(x+y+z, x y, x^{2}+y^{2}+z^{2}\right), g(u, v, w)=\left(\mathrm{e}^{v^{2}+w^{2}}, \sin u w, \sqrt{u v}\right)$, 在
	点 $(u, v, w)=(2,1,3)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\boldsymbol{f}\circ\boldsymbol{g}    & = \left(\begin{array}{c}
				f_1(g_1,g_2,g_3) \\
				f_2(g_1,g_2,g_3) \\
				f_3(g_1,g_2,g_3)
			\end{array}\right)                                            \\
		(\boldsymbol{f}\circ\boldsymbol{g})' & =\left(\begin{array}{c}
				\nabla f_1(g_1,g_2,g_3) \\
				\nabla f_2(g_1,g_2,g_3) \\
				\nabla f_3(g_1,g_2,g_3) \\
			\end{array}\right)\cdot\left(\begin{array}{ccc}
				\dfrac{\partial g_1}{\partial u} & \dfrac{\partial g_1}{\partial v} & \dfrac{\partial g_1}{\partial w} \\
				\dfrac{\partial g_2}{\partial u} & \dfrac{\partial g_2}{\partial v} & \dfrac{\partial g_2}{\partial w} \\
				\dfrac{\partial g_3}{\partial u} & \dfrac{\partial g_3}{\partial v} & \dfrac{\partial g_3}{\partial w} \\
			\end{array}\right) \\
		                                     & =\left(\begin{array}{ccc}
				1  & 1  & 1  \\
				y  & x  & 0  \\
				2x & 2y & 2z
			\end{array}\right)\cdot\left(\begin{array}{ccc}
				0                           & 2ve^{v^2+w^2}               & 2we^{v^2+w^2} \\
				w\cos uw                    & 0                           & u\cos uw      \\
				\dfrac{\sqrt{v}}{2\sqrt{u}} & \dfrac{\sqrt{u}}{2\sqrt{v}} & 0
			\end{array}\right) \\
		                                     & =\left(\begin{array}{ccc}
				w\cos uw + \dfrac{\sqrt{v}}{2\sqrt{u}} & 2ve^{v^2+w^2}+\dfrac{\sqrt{u}}{2\sqrt{v}} & 2we^{v^2+w^2}+u\cos uw                     \\
				we^{v^2+w^2}\cos uw                    & 2ve^{v^2+w^2}\sin uw                      & 2we^{v^2+w^2}\sin uw + ue^{v^2+w^2}\cos uw \\
				2w\cos uw \sin uw + v                  & 4ve^{2v^2+2w^2} + u                       & 4we^{2v^2+2w^2} + 2u\cos uw\sin uw
			\end{array}\right)
	\end{aligned}
$$
将$(u,v,w) = (2,1,3)$带入得
$$
	(\boldsymbol{f}\circ\boldsymbol{g})'=\left(\begin{array}{ccc}
			3\cos 6 + \dfrac{1}{2\sqrt{2}} & 2e^{10}+\dfrac{1}{\sqrt{2}} & 6e^{10}+2\cos 6             \\
			3e^{10}\cos 6                  & 2e^{10}\sin 6               & 6e^{10}\sin 6+2e^{10}\cos 6 \\
			6\cos 6\sin 6+1                & 4e^{20}+2                   & 12e^{20}+4\cos 6\sin 6\end{array}\right)
$$
\section{如果函数 $f(x, y)$ 满足: 对于任意实数 $t$ 及 $x, y$, 成立$$f(t x, t y)=t^{n} f(x, y)$$则称 $f$ 为 $n$ 次齐次函数.证明 $:n$ 次齐次函数满足方程$$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=n f$$}
\begin{proof}
	对等式两边求导:
	$$
		\begin{aligned}
			\dfrac{\mathrm{~d}f(tx,ty)}{\mathrm{~d}t}  & =x\dfrac{\partial f(tx,ty)}{\partial x}+y\dfrac{\partial f(tx,ty)}{\partial y} \\
			\dfrac{\mathrm{~d}t^nf(x,y)}{\mathrm{~d}t} & =nt^{n-1}f(x,y)
		\end{aligned}
	$$
	取$t=1$即证.
\end{proof}

\section{设 $z=f(xy), f$ 为可微的一元函数,证明: $x \dfrac{\partial z}{\partial x}-y \dfrac{\partial z}{\partial y}=0$}
\begin{proof}
	$$
		\begin{aligned}
			\dfrac{\partial z}{\partial x} & =\dfrac{\partial f}{\partial xy}\cdot\dfrac{\partial xy}{\partial x} \\
			                               & =\dfrac{\partial f}{\partial xy}\cdot y                              \\
			\dfrac{\partial z}{\partial x} & =\dfrac{\partial f}{\partial xy}\cdot\dfrac{\partial xy}{\partial y} \\
			                               & =\dfrac{\partial f}{\partial xy}\cdot x
		\end{aligned}
	$$
	将其带入即证.
\end{proof}


\section{若二元函数 $z=f(x,y)$ 满足方程 $x \dfrac{\partial z}{\partial x}+y \dfrac{\partial z}{\partial y}=0 .$ 证明 $:$ 该函数在极坐标系下只是
  $\theta$ 的函数.}
\begin{proof}
	$x = \rho \cos \theta, y = \rho \sin \theta$.从而
	$$
		\begin{aligned}
			\dfrac{\mathrm{~d}z}{\mathrm{~d}\rho} & =\dfrac{\partial z}{\partial x}\cdot\dfrac{\partial x}{\partial \rho}+\dfrac{\partial z}{\partial y}\cdot\dfrac{\partial y}{\partial \rho} \\
			                                      & =\dfrac{\partial z}{\partial x}\cdot\cos \theta+\dfrac{\partial z}{\partial y}\cdot\sin \theta
		\end{aligned}
	$$
	由于$\rho\cdot\dfrac{\partial z}{\partial x}\cdot\cos \theta+\rho\dfrac{\partial z}{\partial y}\cdot\sin \theta=x \dfrac{\partial z}{\partial x}+y \dfrac{\partial z}{\partial y}=0$,从而$\dfrac{\mathrm{~d}z}{\mathrm{~d}\rho}=0$,所以其为关于$\theta$的函数.
\end{proof}
\end{document}